I was recently thinking back to the collatz conjecture and decided to just think about it for a while. For those who do not know, the conjecture is basically this:

If n is odd, multiply n by three then add one. If n is even, divide n by 2. Take that answer and repeat the process such that you have a sequence. For any positive even integer n, does the sequence ever not terminate at 1?

My thoughts:

Any sequence which reaches 1 terminates as it enters the loop 1>4>2>1

Any sequence which contains a number known to terminate at 1, also terminates at one

All odd numbers do not need to be checked

Take a and b to both be positive odd integers.

a can be written as 2c+1 by the definition of an odd number

b can be written as 2d+1 by the definition of an odd number

a*b = (2c+1)(2d+1) by subsitution

(2c+1)(2d+1) = 4cd+2c+2d+1 by distribution

4cd+2c+2d+1 = (4cd+2c+2d)+1 by association

(4cd+2c+2d)+1 = 2(2cd+c+d)+1 by distribution

2(2cd+c+d)+1 is an odd number by definition. any integer multiplied by 2 and added to 1 is odd.

Any odd number plus one is even by definition

Therefore, after one iteration any odd number will become even.

For any integer a where n is equal to 2^a the sequence will terminate at 1

Since 2^a is even, and can be divided by 2 exactly a times, the product of each iteration becomes 2^a-k where k is the current iteration

When a=k then the product of the iteration will be 2^a-a which is 2^0 which is equal to 1

Any sequence which enters a loop of any size is considered terminated.

Any sequence which diverges, or never repeats a number, is considered to grow to infinity

Based on the proof above, does that mean that a solution to the problem would be to determine if all the series eventually reach a point of 2^n? Thereby making the conjecture:

If n is even divide n by 2. If n is odd, multiply n by three and add one, then divide by two. Take that answer and repeat the process such that you have a sequence. For any positive integer n, does the sequence ever not contain a value 2^n?

which can easily shown to yield the same iterations as the original definition. A few other notes:

If you were to represent i iterations of some input n as a function of n, then this function will be identical for all n that are congruent modulo $2^i$. This means that, for example, there is a function that maps all values of n that leave remainder 1 when divided by 4 to the second iteration of the Collatz function for that n.

The calculation of the function is easily shown to be based on the value $2^i$ and the number of odd steps in the iteration.

If any loop other than $4 \rightarrow 2 \rightarrow 1 \rightarrow 4$ exists, it much have at least 68 members.

Heuristic arguments suggest no number can grow indefinitely, as each input is either halved or multiplied by 1.5, though this is not a valid proof.

The Collatz function and its shortcut can also be represented as singular functions involving $\sin$ or $\cos$ due to those functions cyclical behavior.

Most of this info comes directly from the Wikipedia page or papers referenced there.

Gladly. Imagine you want to calculate the third iteration of the "shortcut" Collatz function without calculating any iterations in between for any n. There are $2^3=8$ such functions, which depend solely on $n \pmod 8$

Note that each function is divided by 8 and that the coefficient of n is a power of 3, specifically $3^o$ where o is the number of odd numbers encountered in the iterations. I did a bit of independent research a while back to find a pattern to these functions, but my searches on the OEIS and WolframAlpha were not very fruitful.

Wonderful simplification! Being only in high school I haven't had a chance to study algebraic rings in-depth (my breadth of knowledge is Wikipedia articles).
One thing I did note about my initial result was that $2n+1$ is the derivative of $n^2+n$. No idea if this is in any way significant, but I thought it was cool to note.